Question

A uniform disk of mass 20.0 kg and radius 20.0 cm has an additional rim of mass 20.0 kg as well as four symmetrically placed masses, each of mass 1/4th of the mass of the disk, fastened at positions having position vectors (10.0 i + 10.0 j) cm, (10.0 i - 10.0 j) cm, (-10.0i - 10.0 j cm, (-10.0 i + 10.0 j) cm (with respect to the center of the disk). What is the moment of inertia of the whole unit about an axis perpendicular to the disk and passing through its center?

Answer 1

`I = 1.6 kg m^2`

Step-by-step explanation:

Moment of inertia of disc is given as

`I = (1)/(2)mR^2`

now we have

m = 20 kg

R = 20.0 cm = 0.20 m

now we have

`I_(disc) = (1)/(2)(20 kg)(0.20 m)^2`

`I_(disc) = 0.4 kg m^2`

Now the additional mass of 20 kg is placed on its rim so it will behave as a ring so moment of inertia of that part of the disc is

`I = mR^2`

m = 20 kg

R = 20 cm = 0.20 m

`I_(ring) = 20(0.20^2)`

`I_(ring) = 0.8 kg m^2`

Now four point masses each of the mass of one fourth of mass of disc is placed at four positions so moment of inertia of these four masses is given as

`I_(mass) = 4( m'r^2)`

here we have

`m' = (m)/(4)`

`I_(mass) = 4((m)/(4))(0.10^2 + 0.10^2)`

`I_(mass) = 20(0.02) = 0.40 kg m^2`

Now total moment of inertia of the system is given as

`I = I_(disc) + I_(ring) + I_(mass)`

`I = 0.4 + 0.8 + 0.4 = 1.6 kg m^2`